3.181 \(\int \frac {\tanh ^2(x)}{a+b \cosh (x)} \, dx\)

Optimal. Leaf size=61 \[ \frac {2 \sqrt {a-b} \sqrt {a+b} \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a^2}+\frac {b \tan ^{-1}(\sinh (x))}{a^2}-\frac {\tanh (x)}{a} \]

[Out]

b*arctan(sinh(x))/a^2+2*arctanh((a-b)^(1/2)*tanh(1/2*x)/(a+b)^(1/2))*(a-b)^(1/2)*(a+b)^(1/2)/a^2-tanh(x)/a

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Rubi [A]  time = 0.23, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {2723, 3056, 3001, 3770, 2659, 208} \[ \frac {2 \sqrt {a-b} \sqrt {a+b} \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a^2}+\frac {b \tan ^{-1}(\sinh (x))}{a^2}-\frac {\tanh (x)}{a} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]^2/(a + b*Cosh[x]),x]

[Out]

(b*ArcTan[Sinh[x]])/a^2 + (2*Sqrt[a - b]*Sqrt[a + b]*ArcTanh[(Sqrt[a - b]*Tanh[x/2])/Sqrt[a + b]])/a^2 - Tanh[
x]/a

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 2723

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)/tan[(e_.) + (f_.)*(x_)]^2, x_Symbol] :> Int[((a + b*Sin[e + f*
x])^m*(1 - Sin[e + f*x]^2))/Sin[e + f*x]^2, x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - b^2, 0]

Rule 3001

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] + Dist[(B*c - A
*d)/(b*c - a*d), Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 3056

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c +
d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), I
nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*
(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d*(A*b^2 + a^2*C)*(m + n + 3)
*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
&& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ
[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \frac {\tanh ^2(x)}{a+b \cosh (x)} \, dx &=-\int \frac {\left (1-\cosh ^2(x)\right ) \text {sech}^2(x)}{a+b \cosh (x)} \, dx\\ &=-\frac {\tanh (x)}{a}-\frac {\int \frac {(-b-a \cosh (x)) \text {sech}(x)}{a+b \cosh (x)} \, dx}{a}\\ &=-\frac {\tanh (x)}{a}+\frac {b \int \text {sech}(x) \, dx}{a^2}-\frac {\left (-a^2+b^2\right ) \int \frac {1}{a+b \cosh (x)} \, dx}{a^2}\\ &=\frac {b \tan ^{-1}(\sinh (x))}{a^2}-\frac {\tanh (x)}{a}-\frac {\left (2 \left (-a^2+b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a+b-(a-b) x^2} \, dx,x,\tanh \left (\frac {x}{2}\right )\right )}{a^2}\\ &=\frac {b \tan ^{-1}(\sinh (x))}{a^2}+\frac {2 \sqrt {a-b} \sqrt {a+b} \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a^2}-\frac {\tanh (x)}{a}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 61, normalized size = 1.00 \[ \frac {2 \sqrt {b^2-a^2} \tan ^{-1}\left (\frac {(a-b) \tanh \left (\frac {x}{2}\right )}{\sqrt {b^2-a^2}}\right )-a \tanh (x)+2 b \tan ^{-1}\left (\tanh \left (\frac {x}{2}\right )\right )}{a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[x]^2/(a + b*Cosh[x]),x]

[Out]

(2*b*ArcTan[Tanh[x/2]] + 2*Sqrt[-a^2 + b^2]*ArcTan[((a - b)*Tanh[x/2])/Sqrt[-a^2 + b^2]] - a*Tanh[x])/a^2

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fricas [B]  time = 0.63, size = 326, normalized size = 5.34 \[ \left [\frac {\sqrt {a^{2} - b^{2}} {\left (\cosh \relax (x)^{2} + 2 \, \cosh \relax (x) \sinh \relax (x) + \sinh \relax (x)^{2} + 1\right )} \log \left (\frac {b^{2} \cosh \relax (x)^{2} + b^{2} \sinh \relax (x)^{2} + 2 \, a b \cosh \relax (x) + 2 \, a^{2} - b^{2} + 2 \, {\left (b^{2} \cosh \relax (x) + a b\right )} \sinh \relax (x) - 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cosh \relax (x) + b \sinh \relax (x) + a\right )}}{b \cosh \relax (x)^{2} + b \sinh \relax (x)^{2} + 2 \, a \cosh \relax (x) + 2 \, {\left (b \cosh \relax (x) + a\right )} \sinh \relax (x) + b}\right ) + 2 \, {\left (b \cosh \relax (x)^{2} + 2 \, b \cosh \relax (x) \sinh \relax (x) + b \sinh \relax (x)^{2} + b\right )} \arctan \left (\cosh \relax (x) + \sinh \relax (x)\right ) + 2 \, a}{a^{2} \cosh \relax (x)^{2} + 2 \, a^{2} \cosh \relax (x) \sinh \relax (x) + a^{2} \sinh \relax (x)^{2} + a^{2}}, -\frac {2 \, {\left (\sqrt {-a^{2} + b^{2}} {\left (\cosh \relax (x)^{2} + 2 \, \cosh \relax (x) \sinh \relax (x) + \sinh \relax (x)^{2} + 1\right )} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cosh \relax (x) + b \sinh \relax (x) + a\right )}}{a^{2} - b^{2}}\right ) - {\left (b \cosh \relax (x)^{2} + 2 \, b \cosh \relax (x) \sinh \relax (x) + b \sinh \relax (x)^{2} + b\right )} \arctan \left (\cosh \relax (x) + \sinh \relax (x)\right ) - a\right )}}{a^{2} \cosh \relax (x)^{2} + 2 \, a^{2} \cosh \relax (x) \sinh \relax (x) + a^{2} \sinh \relax (x)^{2} + a^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^2/(a+b*cosh(x)),x, algorithm="fricas")

[Out]

[(sqrt(a^2 - b^2)*(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 + 1)*log((b^2*cosh(x)^2 + b^2*sinh(x)^2 + 2*a*b*c
osh(x) + 2*a^2 - b^2 + 2*(b^2*cosh(x) + a*b)*sinh(x) - 2*sqrt(a^2 - b^2)*(b*cosh(x) + b*sinh(x) + a))/(b*cosh(
x)^2 + b*sinh(x)^2 + 2*a*cosh(x) + 2*(b*cosh(x) + a)*sinh(x) + b)) + 2*(b*cosh(x)^2 + 2*b*cosh(x)*sinh(x) + b*
sinh(x)^2 + b)*arctan(cosh(x) + sinh(x)) + 2*a)/(a^2*cosh(x)^2 + 2*a^2*cosh(x)*sinh(x) + a^2*sinh(x)^2 + a^2),
-2*(sqrt(-a^2 + b^2)*(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 + 1)*arctan(-sqrt(-a^2 + b^2)*(b*cosh(x) + b*
sinh(x) + a)/(a^2 - b^2)) - (b*cosh(x)^2 + 2*b*cosh(x)*sinh(x) + b*sinh(x)^2 + b)*arctan(cosh(x) + sinh(x)) -
a)/(a^2*cosh(x)^2 + 2*a^2*cosh(x)*sinh(x) + a^2*sinh(x)^2 + a^2)]

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giac [A]  time = 0.13, size = 67, normalized size = 1.10 \[ \frac {2 \, b \arctan \left (e^{x}\right )}{a^{2}} + \frac {2 \, {\left (a^{2} - b^{2}\right )} \arctan \left (\frac {b e^{x} + a}{\sqrt {-a^{2} + b^{2}}}\right )}{\sqrt {-a^{2} + b^{2}} a^{2}} + \frac {2}{a {\left (e^{\left (2 \, x\right )} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^2/(a+b*cosh(x)),x, algorithm="giac")

[Out]

2*b*arctan(e^x)/a^2 + 2*(a^2 - b^2)*arctan((b*e^x + a)/sqrt(-a^2 + b^2))/(sqrt(-a^2 + b^2)*a^2) + 2/(a*(e^(2*x
) + 1))

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maple [B]  time = 0.09, size = 108, normalized size = 1.77 \[ \frac {2 \arctanh \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}-\frac {2 \arctanh \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right ) b^{2}}{a^{2} \sqrt {\left (a +b \right ) \left (a -b \right )}}-\frac {2 \tanh \left (\frac {x}{2}\right )}{a \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )}+\frac {2 \arctan \left (\tanh \left (\frac {x}{2}\right )\right ) b}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)^2/(a+b*cosh(x)),x)

[Out]

2/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tanh(1/2*x)/((a+b)*(a-b))^(1/2))-2/a^2/((a+b)*(a-b))^(1/2)*arctanh((a-b)*t
anh(1/2*x)/((a+b)*(a-b))^(1/2))*b^2-2/a*tanh(1/2*x)/(tanh(1/2*x)^2+1)+2/a^2*arctan(tanh(1/2*x))*b

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^2/(a+b*cosh(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?`
for more details)Is 4*a^2-4*b^2 positive or negative?

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mupad [B]  time = 3.53, size = 285, normalized size = 4.67 \[ \frac {2}{a+a\,{\mathrm {e}}^{2\,x}}+\frac {\ln \left (64\,a^3\,b-64\,a\,b^3-32\,b^3\,\sqrt {a^2-b^2}+128\,a^4\,{\mathrm {e}}^x+32\,b^4\,{\mathrm {e}}^x+64\,a^2\,b\,\sqrt {a^2-b^2}+128\,a^3\,{\mathrm {e}}^x\,\sqrt {a^2-b^2}-160\,a^2\,b^2\,{\mathrm {e}}^x-96\,a\,b^2\,{\mathrm {e}}^x\,\sqrt {a^2-b^2}\right )\,\sqrt {a^2-b^2}}{a^2}-\frac {\ln \left (64\,a^3\,b-64\,a\,b^3+32\,b^3\,\sqrt {a^2-b^2}+128\,a^4\,{\mathrm {e}}^x+32\,b^4\,{\mathrm {e}}^x-64\,a^2\,b\,\sqrt {a^2-b^2}-128\,a^3\,{\mathrm {e}}^x\,\sqrt {a^2-b^2}-160\,a^2\,b^2\,{\mathrm {e}}^x+96\,a\,b^2\,{\mathrm {e}}^x\,\sqrt {a^2-b^2}\right )\,\sqrt {a^2-b^2}}{a^2}-\frac {b\,\left (\ln \left ({\mathrm {e}}^x-\mathrm {i}\right )\,1{}\mathrm {i}-\ln \left ({\mathrm {e}}^x+1{}\mathrm {i}\right )\,1{}\mathrm {i}\right )}{a^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)^2/(a + b*cosh(x)),x)

[Out]

2/(a + a*exp(2*x)) + (log(64*a^3*b - 64*a*b^3 - 32*b^3*(a^2 - b^2)^(1/2) + 128*a^4*exp(x) + 32*b^4*exp(x) + 64
*a^2*b*(a^2 - b^2)^(1/2) + 128*a^3*exp(x)*(a^2 - b^2)^(1/2) - 160*a^2*b^2*exp(x) - 96*a*b^2*exp(x)*(a^2 - b^2)
^(1/2))*(a^2 - b^2)^(1/2))/a^2 - (log(64*a^3*b - 64*a*b^3 + 32*b^3*(a^2 - b^2)^(1/2) + 128*a^4*exp(x) + 32*b^4
*exp(x) - 64*a^2*b*(a^2 - b^2)^(1/2) - 128*a^3*exp(x)*(a^2 - b^2)^(1/2) - 160*a^2*b^2*exp(x) + 96*a*b^2*exp(x)
*(a^2 - b^2)^(1/2))*(a^2 - b^2)^(1/2))/a^2 - (b*(log(exp(x) - 1i)*1i - log(exp(x) + 1i)*1i))/a^2

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tanh ^{2}{\relax (x )}}{a + b \cosh {\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)**2/(a+b*cosh(x)),x)

[Out]

Integral(tanh(x)**2/(a + b*cosh(x)), x)

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